Tutorial: Derivatives of logarithmic and exponential functions
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This tutorial: Part B: Derivatives of exponential functions
(This topic is also in Section 4.5 in Applied Calculus or Section 11.5 in Finite Mathematics and Applied Calculus)
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#[Derivative of][Derivada de]# e^{x} #[and][y]# b^{x}
The derivative of $e^x$ and $b^x$ are given by the following formulas
1. Derivative of $e^x$
%%Q Where do these formulas come from?
$\dfrac{d}{dx}(e^x) = e^x$ \gap[20] \t Yes, the derivative of $e^x$ is itself!
2. Derivative of $b^x$
$\dfrac{d}{dx}(b^x) = b^x \ln b$ \gap[1] \t When $b = e$, this agrees with (1); see below.
#[Note][Nota]# When $b = e$ in formula (2), then $\ln b = \ln e = 1$, so (2) gives the same formula as (1).
%%Examples
1. $\dfrac{d}{dx}(5^x) = 5^x\ln\,5$ \gap[5] \t
Some for you
#[Formula][Fórmula]# 2
\\ 2. $\dfrac{d}{dx}(5e^x) = 5e^x$ \gap[5] \t Derivative of a constant times a function
\\ 3. $\dfrac{d}{dx}\left[4(20^x)\right] = 4(20^x)\ln\,20$ \gap[5] \t Derivative of a constant times a function
\\ 4. $\dfrac{d}{dx}\left(x^2e^x\right) = 2xe^x + x^2e^x$ \gap[5] \t Product rule
\\ $\qquad {}= e^x(2x+x^2)$
%%A For derivations of these formulas, consult Section 4.5 in Applied Calculus or Section 11.5 in Finite Mathematics and Applied Calculus.
Derivatives of exponentials of functions
We now know how to differentiate expressions that contain $b^x$ for any base $b$. What about $b$ raised to more complicated quantities, for instance $2^{x^23x+2}$? For things like this we need to use the chain rule (see the %%chainruletut):
Differentiating exponentials of functions
\\ \t
Derivative of $e$ rised to a quantity is $e$ raised to that that quantity times the derivative of that quantity.
\\
\t The derivative of the b raised to a quantity is the product of b raised to that quantity with ln b, times the derivative of that quantity.
\\
%%Examples
5. $\dfrac{d}{dx}\left[e^{\color{blue}{x^2+x}}\right] = e^{\color{blue}{x^2+x}}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \t ${}= (2x+1)e^{x^2+2}$
\\ 6. $\dfrac{d}{dx}\left[3^{\color{blue}{x^2+x}}\right] = \left(3^{\color{blue}{x^2+x}}\ln 3\right)\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \t ${}= (2x+1)3^{x^2+2}\ln 3$
Some for you
Now try the exercises in Section 4.5 in Applied Calculus or Section 11.5 in Finite Mathematics and Applied Calculus.
Copyright © 2019 Stefan Waner and Steven R. Costenoble
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